I am having troubles with deriving the upper bounds on option prices. For example, I am trying to determine the upper bound on a European put option.

If I define the put price as $P$, the terminal stock price as $S_T$, the strike as $X$, the risk free rates as $r$, and the time to expiry as $T$, I am able to create the following portfolio;

Short 1 Put, and invest the proceeds.

So at the outset, from the short put position I gain $P$, and I invest an equal amount at the risk free rate, giving me a net position of $0$.

At expiry, if $S_T > X$, I have zero cash flow from the short put position, and have positive cash flow of $P\text{e}^{rT}$. My net position is $P\text{e}^{rT}$, which is a non negative amount.

If however at expiry $S_T < X$, I have a cash flow of $-(X-S_T)$ from the short position and $P\text{e}^{rT}$ from the invested proceeds. This is a net position of $P\text{e}^{rT} - X + S_T$.

From my understanding, for this to satisfy a no arbitrage condition, we need to have $P\text{e}^{rT} - X + S_T < 0$. Rearranging this should show that the upper bound for the put option needs to be $P < X\text{e}^{rT} - S_T\text{e}^{rT}$.

However, when I consult sources (such as Hull), I am told the upper bound is $P < X\text{e}^{rT}$.

Where is the mistake in my work? This will also help me understand the upper bounds for European call options! Thanks!


2 Answers 2


First, I think you got discounting and compounding wrong. The upper bound for the put price is $P_0 < X e^{-r T}$. This has to hold as long as $S_0 > 0$.

Your approach is generally correct though. Assume that $P_0 = X e^{-r T}$. Then you invest $P_0$ at time $t = 0$ and have $P_0 e^{r T} = X$ at time $t = T$. Now there are three possible states in time $t = T$:

  1. The put option expires worthless. You you have no further obligations and are left with $V_T = X$.
  2. The option expires in-the-money but $S_T > 0$. Then you have to pay out $P_T < X$ from your short put position, leaving you again with a strictly positive cash-flow $V_T > 0$.
  3. We have $S_T = 0$ and thus $P_T = X$. Your net cash-flow is now zero - $V_T = 0$.

As long as $\mathbb{P} \left\{ S_T = 0 \right\} < 1$, this represents a free lottery arbitrage. I.e. you have a zero initial cash-flow $V_0 = 0$, a non-negative terminal cash-flow with probability one $\mathbb{P} \left\{ V_T \geq 0 \right\} = 1$ and a strictly positive terminal cash-flow with a strictly positive probability $\mathbb{P} \left\{ V_T > 0 \right\} > 0$.

But when $\mathbb{P} \left\{ S_T = 0 \right\} = 1$, then any $S_0 \neq 0$ would already represent an arbitrage by itself.


Have you considered implied volatility ? It might be really high for some volatile stocks, so put price might goes to extremely high values due to iv ...

  • 1
    $\begingroup$ This question is about model independent no-arbitrage bounds. I.e. bounds that have to be satisfied in any model an no matter what the unobservable market parameters such as implied volatility are. $P_0 < X e^{-r T}$ holds in the Black-Scholes model even if $\sigma \rightarrow \infty$. $\endgroup$ Mar 11, 2017 at 19:12

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